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anonymous
 2 years ago
Why do some law of sines problems have none or two solutions?
anonymous
 2 years ago
Why do some law of sines problems have none or two solutions?

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anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Because a number canbe positive or negative and still result in the same outcome in a law of sines problem.

Kaederfds
 2 years ago
Best ResponseYou've already chosen the best response.0and it can be has more than 2 solutions too

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0can you explain better please...

Kaederfds
 2 years ago
Best ResponseYou've already chosen the best response.0because sin has periodic

jim_thompson5910
 2 years ago
Best ResponseYou've already chosen the best response.2This page may help you out http://www.regentsprep.org/Regents/math/algtrig/ATT12/lawofsinesAmbiguous.htm

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0because it depends on the quadrant of which the equation puts it. for example, sin is positive in the first and second quadrant (ASTC).. if your number is positive then it would be in quadrants I & II. if your number is negative then it would be in quadrants III & IV

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0jim do you agree with soccergurls answer?

jim_thompson5910
 2 years ago
Best ResponseYou've already chosen the best response.2yes, ATSC is an acronym that refers to you going from quadrant I to quadrant IV A: all are positive in quadrant 1 (sine, cosine, tangent) S: only sine is positive in quadrant 2 T: only tangent is positive in quadrant 3 C: only cosine is positive in quadrant 4 also, sine is positive for quadrants 1 & 2

jim_thompson5910
 2 years ago
Best ResponseYou've already chosen the best response.2and as the page I posted points out, sometimes you'll get something like sin(C) = 1.1428 but sin(x) is restricted from 1 to +1, which means sin(C) = 1.1428 has no solution and therefore, a triangle is not possible

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0ah, i think i get it now, thank you :)

jim_thompson5910
 2 years ago
Best ResponseYou've already chosen the best response.2yw, the page also goes over other examples too (of the 2 other cases)

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0umm can i ask a follow up question...?

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0why does law of cosines not have any ambiguous cases?

jim_thompson5910
 2 years ago
Best ResponseYou've already chosen the best response.2To have an ambiguous case, it means that you aren't sure of what the angles are (if all the sides are known). Let's say we know the lengths of all 3 sides of two triangles. Furthermore, let's say that the corresponding lengths are congruent. If that's true, then we can use the SSS property to show that the two triangles are congruent. Now the question is: is it possible to have 2 congruent triangles but have one set of corresponding angles that are incongruent (of different measures)? The answer is no. The reason is the CPCTC property tells us that if we have congruent triangles, then the corresponding parts will be congruent as well. So the corresponding angles will be congruent, which means knowing all 3 sides leads you to find all 3 angles without any ambiguity or confusion. You can find any angle of a triangle using the law of cosines a^2 = b^2 + c^2  2bc*cos(A) where a,b,c are the sides and A,B,C are the angles (angle A is opposite side a, etc etc). So if you know a,b,c then A will be one fixed value and there won't be any ambiguity. This is true for B,C as well.

jim_thompson5910
 2 years ago
Best ResponseYou've already chosen the best response.2Unfortunately, if you're dealing with SSA, then you may run into ambiguity. This is because knowing two sides and the angle that's not between them doesn't uniquely determine a triangle (think back to geometry). So what this means is that if you were to plug in the sides and angle into a^2 = b^2 + c^2  2bc*cos(A), you would get a quadratic which has the potential to a) have 2 different solutions b) have exactly one solution c) have no solutions at all so the ambiguity still creeps up. But again, this is only possible if working with SSA
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